Let $p(z)=z^2+1$ and $U=\mathbb{C}$ \ $[-i,i]$. Show that there exists a holomorphic function $f$ on $U$ such that $[f(z)]^2=p(z)$ $\forall z \in U$. Further determine the two possible values of $$\oint_{\delta B(0,2)} f(z) \,dz$$ where $B(0,2)$ denote the open ball with centre at the origin and radius $2$. I have managed to show the existence of $f$, but I am unable to solve the last part. Please help. Thanks in advance.

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    $\begingroup$ I would try the substitution $z=1/w$. $\endgroup$ Jul 17 '18 at 9:26
  • $\begingroup$ what representation of $f$ did you get from the first part? The information you have could influence what a proof of the second part will look like $\endgroup$ Jul 17 '18 at 9:26
  • $\begingroup$ I took g(z)=z+i/z-i and showed that g'/g has a primitive h to get f(z)=exp((l(z)) where l(z)=h(z)+ constant.Finally I took f(z)=exp(l(z)/2).(z-i) to prove the result. $\endgroup$
    – Ester
    Jul 17 '18 at 9:33
  • $\begingroup$ @Ester ok. Now, you can write $f(z)$ in terms of $z$ alone by back-substituting $I(z)$ and $h(z)$. Once you have done this, are you familiar with the techniques of contour integration? $\endgroup$ Jul 17 '18 at 9:37
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    $\begingroup$ Here is a standard way of constructing a square root: Fix a point $z_0 \in U$ and fix $c \in \mathbb{C}$ so that $c^2 = p(z_0)$. (Notice that $p$ has no zero on $U$.) Then define $f : U \to \mathbb{C}$ by $$ f(z) = c \exp\left\{ \frac{1}{2} \int_{a}^{z} \frac{p'(\xi)}{p(\xi)} \, d\xi \right\}, $$ where the integral is performed over any nice path from $a$ to $z$. You can check that this is well-defined (i.e. $f(z)$ is uniquely determined regardless of the choice of path of integration) and $f(z)^2 = p(z)$. $\endgroup$ Jul 17 '18 at 10:40

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